Difference between revisions of "Math 22 Integration by Parts and Present Value"

Revision as of 05:56, 18 August 2020

 Let  $u$  and  $v$  be differentiable functions of  $x$ .
 
  $\int udv=uv-\int vdu$

Exercises Use integration by parts to evaluation:

1) $\int \ln xdx$

Solution:
Let $u=\ln x$ , $>du={\frac {1}{x}}dx$
and $dv=dx$ and $v=x$
Then, by integration by parts: $\int \ln xdx=x\ln x-\int x{\frac {1}{x}}dx=x\ln x-\int dx=x\ln x-x+C$

2) $\int xe^{3x}dx$

Solution:
Let $u=x$ , $du=dx$
and $dv=e^{3x}dx$ and $v={\frac {1}{3}}e^{3x}$
Then, by integration by parts: $\int xe^{3x}dx=x{\frac {1}{3}}e^{3x}-\int {\frac {1}{3}}e^{3x}dx=x{\frac {1}{3}}e^{3x}-{\frac {1}{9}}e^{3x}$

This page were made by Tri Phan

@@ Line 5: / Line 5: @@
    <math>\int u dv=uv-\int v du</math>
+'''Exercises''' Use integration by parts to evaluation:
+'''1)''' <math>\int \ln x dx</math>
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+!Solution: &nbsp;
+|-
+|Let <math>u=\ln x</math>, <math>>du=\frac{1}{x}dx</math>
+|-
+|and <math>dv=dx</math> and <math>v=x</math>
+|-
+|Then, by integration by parts: <math>\int \ln x dx=x\ln x-\int x\frac{1}{x}dx=x\ln x-\int dx=x\ln x -x +C</math>
+|}
+'''2)''' <math>\int xe^{3x}dx</math>
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+!Solution: &nbsp;
+|-
+|Let <math>u=x</math>, <math>du=dx</math>
+|-
+|and <math>dv=e^{3x}dx</math> and <math>v=\frac{1}{3}e^{3x}</math>
+|-
+|Then, by integration by parts: <math>\int xe^{3x}dx=x\frac{1}{3}e^{3x} -\int\frac{1}{3}e^{3x} dx=x\frac{1}{3}e^{3x}-\frac{1}{9}e^{3x} </math>
+|}